Consider the ellipse $$\frac{x^{2}}{4}+\frac{y^{2}}{3}=1.$$ Let $H(\alpha, 0), 0<\alpha<2$, be a point. A straight line drawn through $H$ parallel to the $y$-axis crosses the ellipse and its auxiliary circle at points $E$ and $F$ respectively, in the first quadrant. The tangent to the ellipse at the point $E$ intersects the positive $x$-axis at a point $G$. Suppose the straight line joining $F$ and the origin makes an angle $\phi$ with the positive $x$-axis. | List-I | | List-II | | |---|---|---|---| | (I) | If $\phi=\frac{\pi}{4}$, then the area of the triangle $F G H$ is | (P) | $\frac{(\sqrt{3}-1)^{4}}{8}$ | | (II) | If $\phi=\frac{\pi}{3}$, then the area of the triangle $F G H$ is | (Q) | 1 | | (III) | If $\phi=\frac{\pi}{6}$, then the area of the triangle $F G H$ is | (R) | $\frac{3}{4}$ | | (IV) | If $\phi=\frac{\pi}{12}$, then the area of the triangle $F G H$ is | (S) | $\frac{1}{2 \sqrt{3}}$ | | | | (T) | $\frac{3 \sqrt{3}}{2}$ | The correct option is:

(C) (I) → (Q); (II) → (T); (III) → (S); (IV) → (P)

(A) (I) → (R); (II) → (S); (III) → (Q); (IV) → (P)

(B) (I) → (R); (II) → (T); (III) → (S); (IV) → (P)

(D) (I) → (Q); (II) → (S); (III) → (Q); (IV) → (P)

Ellipse — JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 Question with Solution

Consider the ellipse

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1.

Let

H (α, 0), 0 < α < 2

, be a point. A straight line drawn through

H

parallel to the

y

-axis crosses the ellipse and its auxiliary circle at points

E

and

F

respectively, in the first quadrant. The tangent to the ellipse at the point

E

intersects the positive

x

-axis at a point

G

. Suppose the straight line joining

F

and the origin makes an angle

ϕ

with the positive

x

-axis.

List-I		List-II
(I)	If $ϕ = \frac{π}{4}$ , then the area of the triangle $F G H$ is	(P)	$\frac{( 3 - 1 ) ^{4}}{8}$
(II)	If $ϕ = \frac{π}{3}$ , then the area of the triangle $F G H$ is	(Q)	1
(III)	If $ϕ = \frac{π}{6}$ , then the area of the triangle $F G H$ is	(R)	$\frac{3}{4}$
(IV)	If $ϕ = \frac{π}{12}$ , then the area of the triangle $F G H$ is	(S)	$\frac{1}{2 3}$
		(T)	$\frac{3 3}{2}$

The correct option is:

Answer: (C)

The auxiliary circle has equation

x^{2} + y^{2} = 4

. Point

F

on the circle at angle

ϕ

F (2 cos ϕ, 2 sin ϕ)

.

The vertical line through

F

intersects the ellipse at

E (2 cos ϕ, 3 sin ϕ)

.

Tangent to ellipse at

E

\frac{x \cdot 2 c o s ϕ}{4} + \frac{y \cdot 3 s i n ϕ}{3} = 1

, which gives

G (\frac{2}{c o s ϕ}, 0)

.

Also,

H (α, 0) = (2 cos ϕ, 0)

.

Area of triangle

F G H

Area = \frac{1}{2} \cdot H G \cdot F H = \frac{1}{2} (\frac{2}{cos ϕ} - 2 cos ϕ) \cdot 2 sin ϕ = 2 tan ϕ sin^{2} ϕ

Checking:

(I) $ϕ = π /4$ : Area $= 2 \cdot 1 \cdot (1/2) = 1$ → (Q)
(II) $ϕ = π /3$ : Area $= 23 \cdot (3/4) = 33 /2$ → (T)
(III) $ϕ = π /6$ : Area $= 2 \cdot (1/ 3) \cdot (1/4) = 1/ (23)$ → (S)
(IV) $ϕ = π /12$ : Area $= 2 (2 - 3) \cdot (3 - 1)^{2} /8 = (3 - 1)^{4} /8$ → (P)

Correct matching: (I)→(Q), (II)→(T), (III)→(S), (IV)→(P), which is option (C).

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Single correct option